$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is trebled at the instant that it is at a distance $\frac{2A}{3}$ from the equilibrium position. The new amplitude of the motion is

$A$ particle executes $S.H.M.$ with amplitude $a$ and time period $T$. The displacement of the particle when its speed is half of its maximum speed is $\frac{\sqrt{x} a}{2}$. The value of $x$ is $\ldots \ldots \ldots$

$A$ particle is executing simple harmonic motion with amplitude $A$. At a distance $x$ from the mean position,when the particle is moving towards the extreme position,it receives a blow in the direction of motion which instantaneously doubles its velocity. What is the new amplitude of the particle? (Frequency remains constant during the motion)

$A$ particle is executing $S.H.M.$ and its velocity $v$ is related to its position $x$ as $v^2 + ax^2 = b$,where $a$ and $b$ are positive constants. The frequency of oscillation of the particle is ..........

The velocity at the mean position of a particle executing $S.H.M.$ is $v$. What is the velocity of the particle at a distance equal to half of the amplitude?

The amplitude of a particle executing $SHM$ is $4 \,cm$. At the mean position,the speed of the particle is $16 \,cm/s$. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3} \,cm/s$ will be .... $cm$.